That's not as obvious as the usage of the word "just" seems to imply. When we start with something like i^i, even before raising it to another power of i, we first need to understand i^i. The rules for exponentiation that hold good for real numbers cannot be apply here.
How exactly is i^i defined? What does raising a complex number to the power of another complex number even mean? We define it! We first define the following: For complex numbers w and z, w^z = e^(z log w).
Now we use this definition to see what i^i is. We get i^i = e^(i log i) = e^(i * (2i * pi * n + i * pi /2)) = e^(-2 * pi * n - pi/2) for n ∈ ℤ. Note that this is the result of log(i) being multivalued.
So far we have established an interesting result that i^i is always a real number regardless of which value of log(i) we choose from. If we choose the principal value of log(i), i.e., log(i) = i * pi / 2, then i^i = e^(-pi / 2). But let us move on with the multivalued i^i.
We use the result of (i^i) and the definition of w^z to see what (i^i)^i is. We get (i^i)^i = e^(i log(i^i)) = e^(i(2i * pi * m - 2 * pi * n - pi/2)) = e^(-2 * pi * m - 2i * pi * n - i * pi/2)) = e^(-2 * pi * m - i * pi/2)) for m ∈ ℤ.
So we can see that (i^i)^i = -i holds good for a single value of m, i.e., m = 0.
That's not as obvious as the usage of the word "just" seems to imply. When we start with something like i^i, even before raising it to another power of i, we first need to understand i^i. The rules for exponentiation that hold good for real numbers cannot be apply here.
How exactly is i^i defined? What does raising a complex number to the power of another complex number even mean? We define it! We first define the following: For complex numbers w and z, w^z = e^(z log w).
Now we use this definition to see what i^i is. We get i^i = e^(i log i) = e^(i * (2i * pi * n + i * pi /2)) = e^(-2 * pi * n - pi/2) for n ∈ ℤ. Note that this is the result of log(i) being multivalued.
So far we have established an interesting result that i^i is always a real number regardless of which value of log(i) we choose from. If we choose the principal value of log(i), i.e., log(i) = i * pi / 2, then i^i = e^(-pi / 2). But let us move on with the multivalued i^i.
We use the result of (i^i) and the definition of w^z to see what (i^i)^i is. We get (i^i)^i = e^(i log(i^i)) = e^(i(2i * pi * m - 2 * pi * n - pi/2)) = e^(-2 * pi * m - 2i * pi * n - i * pi/2)) = e^(-2 * pi * m - i * pi/2)) for m ∈ ℤ.
So we can see that (i^i)^i = -i holds good for a single value of m, i.e., m = 0.